Mathematics questions and answers

Mathematics Questions and Answers

How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.

436.

Differentiate \(\frac{2x}{\sin x}\) with respect to x.

A.

\(2 \cot x \sec x (1 + \tan x)\)

B.

\(2 \csc x - x \cot x\)

C.

\(2x \csc x + \tan x\)

D.

\(2\csc x(1 - x\cot x)\)

Correct answer is D

No explanation has been provided for this answer.

437.

If y = 8x\(^3\) - 3x\(^2\) + 7x - 1, find \(\frac{\mathrm d^2 y}{\mathrm d x^2}\).

A.

48x - 6

B.

11x\(^2\) + 6x - 7

C.

32x + 7

D.

24x\(^2\) - 6x + 7

Correct answer is A

y = 8x\(^3\) - 3x\(^2\) + 7x - 1

\(\frac{\mathrm d^2 y}{\mathrm d x^2}  = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\)

= \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\)

\(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)

438.

Given the matrix \(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\). Find the inverse of matrix A.

A.

\(\begin{vmatrix} 6 & 2 \\ 1 & 6 \end{vmatrix}\)

B.

\(\begin{vmatrix} \frac{2}{11} & \frac{1}{12}\\ \frac{3}{20} & \frac{1}{10} \end{vmatrix}\)

C.

\(\begin{vmatrix} -3 & 2 \\ -1 & -6 \end{vmatrix}\)

D.

\(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20}\end{vmatrix}\)

Correct answer is D

\(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\)

|A| = (3 x 6) - (-2 x 1)

= 18 + 2

= 20.

A\(^{-1}\) = \(\frac{1}{20} \begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}\)

= \(\begin{vmatrix} \frac{6}{20} & \frac{2}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)

= \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)

439.

If \(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\), find the value of x.

A.

5

B.

8

C.

6

D.

3

Correct answer is B

\(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\)

\(\implies 2 \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - (-5) \begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} + 3 \begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 132\)

\(2(2 - 12) + 5(2x) + 3(3x) = 132\)

\(-20 + 10x + 9x = 132\)

\(19x = 152\)

\(x = 8\)

440.

If 2x\(^2\) + x - 3 divides x - 2, find the remainder.

A.

7

B.

3

C.

5

D.

6

Correct answer is A

When you divide a polynomial p(x) by (x - a), the remainder = p(a)

i.e. In the case of 2x\(^2\) + x - 3 \(\div\) (x - 2), the remainder = p(2).

= 2(2)\(^2\) + 2 - 3

= 8 + 2 - 3

= 7.